Fb = m*g = 4.58kg * 9.8N/kg = 44.9 N. =
Force of the block.
Fn = 44.9 - 10.5*sin25.5 = 40.4 N. =
Normal = Force perpendicular to the floor.
Fk = u*Fn = 0.09 * 40.4 = 3.63 N. = Force of kinetic friction.
a = (Fx-Fk)/m = (10.5*cos25.5-3.63)/4.58
= 5.85 m/s^2.
V = a*t = 5.85 * 5.90 = 34.5 m/s.
d = (V^2-Vo^2)/2a = 34.5^2-0)/11.7 = 101.7 m.
A 4.58 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.5 N at an angle theta = 25.5° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.09. What distance does the block travel in a time of 5.90 s after it starts moving?
2 answers
The answer was 22.2m...